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zavuch27 [327]
3 years ago
5

The partial pressure of argon in the atmosphere is 0.00934 atm. Calculate the partial pressure in mmHg andtorr . Round each of y

our answers to significant digits.
Chemistry
1 answer:
sergey [27]3 years ago
3 0

Answer:

1. 7.0984 mmHg

2. 7.0984 torr

Explanation:

Data obtained from the question include:

Partial Pressure of Ar = 0.00934 atm

1. The value of pressure in mmHg is obtained as shown below:

1 atm = 760mmHg

Therefore, 0.00934 atm = 0.00934 x 760 = 7.0984mmHg

2. The value of the pressure in torr is obtained as shown below:

1 atm = 760torr

Therefore, 0.00934 atm = 0.00934 x 760 = 7.0984torr

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Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of
pantera1 [17]

Answer:

W=5743.1077\ J

Explanation:

The expression for the work done is:

W=RT \ln \left( \dfrac{P_1}{P_2} \right)

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

T is the temperature

P₁ is the initial pressure

P₂ is the final pressure

Given that:

T = 300 K

P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)

W=300\times \:8.314\ln \left(10\right)

W=2.30258\times \:2494.2

W=5743.1077\ J

5 0
3 years ago
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
3 years ago
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